Answer
$[H_3O^+] = [OH^-] = 2.345 \times 10^{-7}M$
Work Step by Step
Neutral solution: $[H_3O^+] = [OH^-]$
So, we can use a unknown named "x", that has this value:
$[H_3O^+] = [OH^-] = x$
Now, we can use the "Kw" equation:
$[H_3O^+] \times [OH^-] = K_w$
$x \times x = 5.5 \times 10^{-14}$
$x^2 = 5.5 \times 10^{-14}$
$x = \sqrt {5.5 \times 10^{-14}}$
$x = 2.345 \times 10^{-7}$
So:
$[H_3O^+] = [OH^-] = 2.345 \times 10^{-7}M$
